A Practical Buffer Problem

 

We want to make 1 liter of 0.4 M acetate buffer at pH 5.22.

 1. apply Henderson-Hasselbach Equation:

 pH = pKa + log (A^-)/(HA) ;  pKa for acetic acid is 4.74

 \  5.22=4.74 + log (A^-)/(HA)

 or   0.48 = log (A^-)/(HA)

 

• taking antilogs: 3 = (A^-)/(HA)                This shows the ratio of base to acid for pH 5.22
• alternatively 3 (HA) = (A^-) or 3(HA) - (A^-)= 0

 

2.  The problem also requires buffer strength be 0.4 M that is :

(A^-) + (HA) = 0.4.

 

  3. We now have 2 equations in 2 unknowns and can solve by simple algebra.

 

3 (HA) - (A-) = 0

1 (HA) + (A-) = 0.4

________________________

4 (HA)      = 0.4  \  (HA) = 0.1 M

 

substituting this value in either equality above gives

         (A^-) = 0.3 M

 

 4.  Since we want 1 liter of buffer we need (moles= M x v)

 

0.1  moles acetic acid and 0.3 moles acetate

 0.1 moles = 6.0 grams acetic acid (MW = 60) and

 

if we used Na Acetate (MW = 82), we need 24.6 grams